Question: Complete the square to solve for $x$. $x^{2}+4x-32 = 0$
Answer: Begin by moving the constant term to the right side of the equation. $x^2 + 4x = 32$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $4$ , half of it would be $2$ , and squaring it gives us ${4}$ $x^2 + 4x { + 4} = 32 { + 4}$ We can now rewrite the left side of the equation as a squared term. $( x + 2 )^2 = 36$ Take the square root of both sides. $x + 2 = \pm6$ Isolate $x$ to find the solution(s). $x = -2\pm6$ So the solutions are: $x = 4 \text{ or } x = -8$ We already found the completed square: $( x + 2 )^2 = 36$